Evaluate: tan^{-1} left( frac{a - b}{1 + ab} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We use the standard identity for the difference of inverse tangents: $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$.A

Vedclass · Accepted Answer

$	an^{-1} a - 	an^{-1} c$

Vedclass · Answer

(B) We use the standard identity for the difference of inverse tangents: $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$.Applying this to the given expression:$	an^{-1} \left( \frac{a - b}{1 + ab} ight) = 	an^{-1} a - 	an^{-1} b$Similarly:$	an^{-1} \left( \frac{b - c}{1 + bc} ight) = 	an^{-1} b - 	an^{-1} c$Adding these two results together:$(	an^{-1} a - 	an^{-1} b) + (	an^{-1} b - 	an^{-1} c)$The term $-	an^{-1} b$ and $+	an^{-1} b$ cancel each other out:$= 	an^{-1} a - 	an^{-1} c$.

Evaluate: $\tan^{-1} \left( \frac{a - b}{1 + ab} \right) + \tan^{-1} \left( \frac{b - c}{1 + bc} \right)$

Similar Questions

The value of $2 \cot ^{-1} \frac{1}{2} - \cot ^{-1} \frac{4}{3}$ is

$\tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) = $

If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$,then $\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)$ is equal to

The value of $\tan \left(2 \tan ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)\right)$ is equal to:

If $0 < x < 1$,then $\sqrt{1+x^2} [\{x \cos (\cot ^{-1} x)+\sin (\cot ^{-1} x)\}^2-1]^{\frac{1}{2}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module